3.100.17 \(\int \frac {-15+\frac {3 e^{1+\frac {e}{5-x}}}{5-x}+33 x-6 x^2}{-5+x} \, dx\)

Optimal. Leaf size=22 \[ 3 \left (-e^{\frac {e}{5-x}}+x-x^2\right ) \]

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Rubi [A]  time = 0.14, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {6688, 2230, 2209} \begin {gather*} -3 x^2+3 x-3 e^{\frac {e}{5-x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-15 + (3*E^(1 + E/(5 - x)))/(5 - x) + 33*x - 6*x^2)/(-5 + x),x]

[Out]

-3*E^(E/(5 - x)) + 3*x - 3*x^2

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2230

Int[(F_)^((e_.) + ((f_.)*((a_.) + (b_.)*(x_)))/((c_.) + (d_.)*(x_)))*((g_.) + (h_.)*(x_))^(m_.), x_Symbol] :>
Int[(g + h*x)^m*F^((d*e + b*f)/d - (f*(b*c - a*d))/(d*(c + d*x))), x] /; FreeQ[{F, a, b, c, d, e, f, g, h, m},
x] && NeQ[b*c - a*d, 0] && EqQ[d*g - c*h, 0]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (3-\frac {3 e^{\frac {-5-e+x}{-5+x}}}{(-5+x)^2}-6 x\right ) \, dx\\ &=3 x-3 x^2-3 \int \frac {e^{\frac {-5-e+x}{-5+x}}}{(-5+x)^2} \, dx\\ &=3 x-3 x^2-3 \int \frac {e^{1-\frac {e}{-5+x}}}{(-5+x)^2} \, dx\\ &=-3 e^{\frac {e}{5-x}}+3 x-3 x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 19, normalized size = 0.86 \begin {gather*} -3 \left (e^{\frac {e}{5-x}}+(-1+x) x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-15 + (3*E^(1 + E/(5 - x)))/(5 - x) + 33*x - 6*x^2)/(-5 + x),x]

[Out]

-3*(E^(E/(5 - x)) + (-1 + x)*x)

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fricas [A]  time = 1.76, size = 45, normalized size = 2.05 \begin {gather*} -3 \, {\left ({\left (x^{2} - x\right )} e - {\left (x - 5\right )} e^{\left (-\frac {{\left (x - 5\right )} \log \left (-x + 5\right ) - x + e + 5}{x - 5}\right )}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*exp(-log(5-x)+1)*exp(exp(-log(5-x)+1))-6*x^2+33*x-15)/(x-5),x, algorithm="fricas")

[Out]

-3*((x^2 - x)*e - (x - 5)*e^(-((x - 5)*log(-x + 5) - x + e + 5)/(x - 5)))*e^(-1)

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giac [A]  time = 0.17, size = 39, normalized size = 1.77 \begin {gather*} -3 \, {\left (x - 5\right )}^{2} {\left (\frac {9 \, e^{3}}{x - 5} + \frac {e^{\left (-\frac {e}{x - 5} + 3\right )}}{{\left (x - 5\right )}^{2}} + e^{3}\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*exp(-log(5-x)+1)*exp(exp(-log(5-x)+1))-6*x^2+33*x-15)/(x-5),x, algorithm="giac")

[Out]

-3*(x - 5)^2*(9*e^3/(x - 5) + e^(-e/(x - 5) + 3)/(x - 5)^2 + e^3)*e^(-3)

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maple [A]  time = 0.13, size = 22, normalized size = 1.00

method	result	size
risch	\(-3 x^{2}+3 x -3 \,{\mathrm e}^{-\frac {{\mathrm e}}{x -5}}\)	\(22\)
default	\(3 x -3 \,{\mathrm e}^{-1} {\mathrm e}^{1-\frac {{\mathrm e}}{x -5}}-3 x^{2}\)	\(28\)
norman	\(\frac {18 x^{2}-3 x^{3}-3 x \,{\mathrm e}^{\frac {{\mathrm e}}{5-x}}+15 \,{\mathrm e}^{\frac {{\mathrm e}}{5-x}}-75}{x -5}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*exp(-ln(5-x)+1)*exp(exp(-ln(5-x)+1))-6*x^2+33*x-15)/(x-5),x,method=_RETURNVERBOSE)

[Out]

-3*x^2+3*x-3*exp(-exp(1)/(x-5))

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maxima [A]  time = 0.35, size = 21, normalized size = 0.95 \begin {gather*} -3 \, x^{2} + 3 \, x - 3 \, e^{\left (-\frac {e}{x - 5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*exp(-log(5-x)+1)*exp(exp(-log(5-x)+1))-6*x^2+33*x-15)/(x-5),x, algorithm="maxima")

[Out]

-3*x^2 + 3*x - 3*e^(-e/(x - 5))

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mupad [B]  time = 0.21, size = 21, normalized size = 0.95 \begin {gather*} 3\,x-3\,{\mathrm {e}}^{-\frac {\mathrm {e}}{x-5}}-3\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((33*x + 3*exp(exp(1 - log(5 - x)))*exp(1 - log(5 - x)) - 6*x^2 - 15)/(x - 5),x)

[Out]

3*x - 3*exp(-exp(1)/(x - 5)) - 3*x^2

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sympy [A]  time = 0.20, size = 17, normalized size = 0.77 \begin {gather*} - 3 x^{2} + 3 x - 3 e^{\frac {e}{5 - x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*exp(-ln(5-x)+1)*exp(exp(-ln(5-x)+1))-6*x**2+33*x-15)/(x-5),x)

[Out]

-3*x**2 + 3*x - 3*exp(E/(5 - x))

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